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 November 1st, 2013, 10:08 AM #1 Newbie   Joined: Nov 2013 Posts: 8 Thanks: 0 Probability help Hi, I have this math problem that I tried to solve using combinations/factorial and so on but with no luck. There are 20 kids at the party. Waiter has exactly 10 hamburgers and 10 cheeseburgers (1 burger for each kid). He flips coin to know what kind of burger (ham or cheese) to give to first kid. Then he repeats this 17 more times (for next 17 kids). When he comes to 19th kid, he doesn't need to flip coin because there are exactly 2 cheeseburgers left. What is the probability that this situation will happen again? So let N be the number of kids, including 2 celebrating kids. What is the probability that these 2 celebrating kids will be left with same type of burger (both get ham or cheese one)? I guess the order matters, so let the 2 kids be always at the end. I have some example results but I can't get them in any way: if there are 6 kids: 0,625 10 kids: 0,7266 256 kids: 0,95 Thanks November 1st, 2013, 10:42 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Probability help The question I have is this: does the waiter have to flip a coin for each of the first 18 kids, or is it possible that the waiter is done flipping coins by the time he gets to the 11th kid (because the first 10 kids all received the same type of burger)? If this happens, does this qualify as the event happening again? November 1st, 2013, 10:55 AM #3 Newbie   Joined: Nov 2013 Posts: 8 Thanks: 0 Re: Probability help I don't have this information specified, but let's say that yes it is qualified as the event happening again. The head of coin is hamburger, and tail is cheeseburger. If head falls 10 times before he reaches the 19th kid, he doesn't flip coin anymore, because no matter what falls, he has only one kind of burger left. November 1st, 2013, 11:09 AM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Probability help There are two different problems here, as icemanfan has identified: 1) The probability that the last two kids get the same burger. 2) The probability that the waiter has to keep tossing the coin until the 18th time and then the last two kids get the same burger. There are more options in 1) than 2), so a greater probability. The answer for 1 is: This is the probability that from the first n-2 events, one burger is picked n/2 times. For n = 20, p_1 is about 1/3. The probability for 2 is slighty different. To get this, you need n-1 burgers of one sort from the first n-3 events and then by a toss of the coin that burger again. The probability of this is: For n = 20, p_2 is about 1/5. November 1st, 2013, 11:33 AM #5 Newbie   Joined: Nov 2013 Posts: 8 Thanks: 0 Re: Probability help Thanks a lot, I know the question is not straightforward, so I am trying to work with the information I have. If the example results I have are wrong, then it's even more dificult to determine which solution is right. What exactly does the 1/2^(n-3) mean in the formula? November 1st, 2013, 11:53 AM #6 Newbie   Joined: Nov 2013 Posts: 8 Thanks: 0 Re: Probability help OK I found it out, the second formula is right for what I need, but it needs to be subtracted from 1 (p = 1 - p_2) November 1st, 2013, 09:18 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Probability help That one's sure a lotta fun... k = probability that 1st 17 are 9 h's and 8 c's probability that last 2 are 2 c's = k/3 Could be worded this way: 20 cards labeled 1 to 20 are shuffled, then turned over one by one. What's the prob. that the 18th < 11 and the last 2 are both > 10. ...ya, ya...big deal  November 2nd, 2013, 12:51 AM   #8
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Re: Probability help

Quote:
 Originally Posted by Denis Could be worded this way: 20 cards labeled 1 to 20 are shuffled, then turned over one by one. What's the prob. that the 18th < 11 and the last 2 are both > 10.
That's the problem where the burgers are chosen at random - so that the probability of getting one burger increases or decreases relative to the other depending on how many of one type have already been picked. If you toss a coin, it's binomial, as you always have a 1/2 chance of getting each burger - until there are none left.

So, with three burgers left, you have a 1/3 chance of getting the odd burger.

But, if you toss a coin, with three burgers left you have a 1/2 chance of getting the odd burger. November 2nd, 2013, 12:36 PM #9 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: Probability help Merci, Pero; finally understood your stuff! Probabilities for even numbered groups 4 to 20: FOR n = 4 TO 20 STEP 2 u = n! v = ((n/2)!)^2 w = (n-4)! x = (((n-4)/2)!)^2 p = (w/x) / (u/v) PRINT n,p ENDLOOP OUTPUT: n , p 4 , 1/6 (of course!) 6 , 1/10 8 , 3/35 10 , 5/63 12 , 5/66 14 , 21/286 16 , 14/195 18 , 6/85 20 , 45/646 (the OP's problem) November 3rd, 2013, 09:35 AM   #10
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Re: Probability help

Quote:
 Originally Posted by Pero There are two different problems here, as icemanfan has identified: 1) The probability that the last two kids get the same burger. 2) The probability that the waiter has to keep tossing the coin until the 18th time and then the last two kids get the same burger. There are more options in 1) than 2), so a greater probability. The answer for 1 is: This is the probability that from the first n-2 events, one burger is picked n/2 times. For n = 20, p_1 is about 1/3. The probability for 2 is slighty different. To get this, you need n-1 burgers of one sort from the first n-3 events and then by a toss of the coin that burger again. The probability of this is: For n = 20, p_2 is about 1/5.
No objection to the formula for 2), but on 1) I think you're forgetting the fact that if 10 burgers of one type are picked before the 18th event, the probabilities for the following events before 19-20 are also 1. Because there are no burgers of that type left, there's only one burger type left for the following kids, and as such, there's no reason to flip the coin again. There's only one way the last two kids don't get the same burger, and that is only possible if the first 18 events are 9 burgers of the first type and 9 burgers of the other type.

Rather, I believe the probability of the last two kids getting the same type is:
Code:
P(the last two kids get the same burger) = 1 - (((1/2)^18) * C(18, 9)) = 0.8145... = 81.45% = about 8/10
...which does seem more reasonable.

Assuming that the amount of burgers for each type is half the number of kids, N, the formula for 1) is:
Code:
P(the last two kids get the same burger) = 1 - (((1/2)^(N - 2)) * C((N - 2), ((N/2) - 1))) Tags probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hbonstrom Applied Math 0 November 17th, 2012 07:11 PM token22 Advanced Statistics 2 April 26th, 2012 03:28 PM naspek Calculus 1 December 15th, 2009 01:18 PM

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